Question

Question 10
A circle has radius $\sqrt{2}$ cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in the major segment is ${45}^{\circ }$.

Answer 1

Draw a circle having centre O. Let AB = 2cm be the chord of a circle. The chord AB is divided by the line OM into two equal segments.


To prove that $\angle APB={45}^{\circ }$
Here, AN = NB = 1cm
And OB = $\sqrt{2}$cm
$In\Delta ONB,O{B}^2=O{N}^2+N{B}^2$ [use Pythagoras theorem]
$\Rightarrow (\sqrt{2}{)}^2=O{N}^2+(1{)}^2$
$\Rightarrow O{N}^2=2-1=1$
ON = 1 cm [taking positive square root, because distance is always positive]

Also, $\angle ONB={90}^{\circ }$ [ON is the perpendicular bisector of the chord AB]
$\therefore \angle NOB=\angle NBO={45}^{\circ }$ [ON =NB =1cm,angles opposite to equal sides are also equal]
Similarly, $\angle AON={45}^{\circ }$
Now, $\angle AOB=\angle AON+\angle NOB$
$={45}^{\circ }+{45}^{\circ }={90}^{\circ }$
We know that, chord subtends an angle to the circle is half the angle subtended by it to the centre.
$\therefore \angle APB=\frac{1}{2}\angle AOB$
$\angle APB=\frac{{90}^{\circ }}{2}={45}^{\circ }$, hence proved.